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JEE Main 20203 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
3 September
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2020 · 3 September · Evening Shift · Chemistry · Chemical Kinetics

For the reaction 2A + 3B + (3)/(2)C → 3P , which statement is...

For the reaction 2A+3B+32C→3P2A + 3B + \dfrac{3}{2}C \to 3P, which statement is correct?

Options

  1. AdnAdt=32dnBdt=34dnCdt\dfrac{d{n}_{A}}{dt} = \dfrac{3}{2}\dfrac{d{n}_{B}}{dt} = \dfrac{3}{4}\dfrac{d{n}_{C}}{dt}
  2. BdnAdt=dnBdt=dnCdt\dfrac{{dn}_{A}}{dt} = \dfrac{{dn}_{B}}{dt} = \dfrac{{dn}_{C}}{dt}
  3. CdnAdt=23dnBdt=43dnCdt\dfrac{{dn}_{A}}{dt} = \dfrac{2}{3}\dfrac{{dn}_{B}}{dt} = \dfrac{4}{3}\dfrac{{dn}_{C}}{dt}
  4. DdnAdt=23dnBdt=34dnCdt\dfrac{{dn}_{A}}{dt} = \dfrac{2}{3}\dfrac{{dn}_{B}}{dt} = \dfrac{3}{4}\dfrac{{dn}_{C}}{dt}

Correct answer

C. dnAdt=23dnBdt=43dnCdt\dfrac{{dn}_{A}}{dt} = \dfrac{2}{3}\dfrac{{dn}_{B}}{dt} = \dfrac{4}{3}\dfrac{{dn}_{C}}{dt}

Step-by-step solution

For a given reaction, rate =−12dnAdt=−13dnBdt=−23dncdt= - \dfrac{1}{2}\dfrac{{dn}_{A}}{dt} = - \dfrac{1}{3}\dfrac{{dn}_{B}}{dt} = - \dfrac{2}{3}\dfrac{{dn}_{c}}{dt}

rate =dnAdt=23dnBdt=43dnCdt= \dfrac{{dn}_{A}}{dt} = \dfrac{2}{3}\dfrac{{dn}_{B}}{dt} = \dfrac{4}{3}\dfrac{{dn}_{C}}{dt}

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